{
  "paper": {
    "school": "Nan Hua Primary School",
    "year": 2025,
    "level": "P6",
    "label": "WA2",
    "source_prefix": "Nan Hua Primary School 2025 P6 WA2",
    "has_answer_key": true
  },
  "questions": [
    {
      "n": 1,
      "type_id": 1,
      "question": "How many hundredths are there in 0.8?",
      "answer0": "0.08",
      "answer1": "0.8",
      "answer2": "8",
      "answer3": "80",
      "correct_answer": 3,
      "skill_id": 123,
      "difficulty_id": 1,
      "explanation": "0.8 = 0.80 = 80 hundredths, since each hundredth is 0.01 and \\(0.80 \\div 0.01 = 80\\).",
      "hints": ["Write 0.8 as 0.80.", "Count how many 0.01s make 0.80."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q1",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 4 (80)."
    },
    {
      "n": 2,
      "type_id": 1,
      "question": "What is the sum of all the factors of 9?",
      "answer0": "12",
      "answer1": "13",
      "answer2": "15",
      "answer3": "16",
      "correct_answer": 1,
      "skill_id": 111,
      "difficulty_id": 1,
      "explanation": "Factors of 9 are 1, 3 and 9. Sum = \\(1 + 3 + 9 = 13\\).",
      "hints": ["List all factors of 9.", "Add 1 + 3 + 9."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q2",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 2 (13)."
    },
    {
      "n": 3,
      "type_id": 1,
      "question": "Express 8 km 20 m in km.",
      "answer0": "8020 m",
      "answer1": "8.002 km",
      "answer2": "8.02 km",
      "answer3": "8.2 km",
      "correct_answer": 2,
      "skill_id": 184,
      "difficulty_id": 2,
      "explanation": "20 m = 0.02 km. So 8 km 20 m = \\(8 + 0.02 = 8.02\\) km.",
      "hints": ["1 km = 1000 m, so 20 m = 0.02 km.", "Add 8 km and 0.02 km."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q3",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 3 (8.02 km)."
    },
    {
      "n": 4,
      "type_id": 1,
      "question": "Express \\(\\dfrac{3}{8}\\) as a decimal correct to 2 decimal places.",
      "answer0": "0.308",
      "answer1": "0.38",
      "answer2": "3.08",
      "answer3": "3.8",
      "correct_answer": 1,
      "skill_id": 158,
      "difficulty_id": 2,
      "explanation": "\\(3 \\div 8 = 0.375\\). Rounded to 2 decimal places, 0.375 ≈ 0.38.",
      "hints": ["Divide 3 by 8.", "Round 0.375 to 2 decimal places."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q4",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 2 (0.38)."
    },
    {
      "n": 5,
      "type_id": 1,
      "question": "What is the smallest number of squares that must be shaded so that the line PQ becomes a line of symmetry?",
      "answer0": "5",
      "answer1": "2",
      "answer2": "3",
      "answer3": "4",
      "correct_answer": 0,
      "skill_id": 145,
      "difficulty_id": 2,
      "explanation": "Each shaded square must have its mirror image across line PQ also shaded. Counting the squares whose reflections are currently unshaded gives 5 squares that must be added.",
      "hints": ["Reflect each shaded square across the diagonal line PQ.", "Count shaded squares whose mirror image is still empty."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q5",
      "image_needed": true,
      "image_options": false,
      "image_file": "q5.png",
      "image_page": 2,
      "image_bbox": [0.15, 0.4, 0.5, 0.66],
      "image_loc": "grid of squares with diagonal line PQ and some shaded squares, middle of page",
      "notes": "Answer key: option 1 (5). Figure (shaded grid with diagonal PQ) needed."
    },
    {
      "n": 6,
      "type_id": 1,
      "question": "A movie started at 10.35 p.m. and ended at 1.15 a.m. How long did the movie last?",
      "answer0": "2 h 10 min",
      "answer1": "2 h 20 min",
      "answer2": "2 h 40 min",
      "answer3": "2 h 50 min",
      "correct_answer": 2,
      "skill_id": 135,
      "difficulty_id": 2,
      "explanation": "From 10.35 p.m. to 12.35 a.m. is 2 h. From 12.35 a.m. to 1.15 a.m. is 40 min. Total = 2 h 40 min.",
      "hints": ["Count whole hours first, then the extra minutes.", "10.35 p.m. → 12.35 a.m. is 2 h."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q6",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 3 (2 h 40 min)."
    },
    {
      "n": 7,
      "type_id": 1,
      "question": "Which one of the following is not a net of the cube?",
      "answer0": "Net 1",
      "answer1": "Net 2",
      "answer2": "Net 3",
      "answer3": "Net 4",
      "correct_answer": 3,
      "skill_id": 233,
      "difficulty_id": 2,
      "explanation": "Folding each arrangement of squares, options 1, 2 and 3 fold into a cube. Option 4 has an arrangement where two faces overlap, so it is not a valid net of a cube.",
      "hints": ["A valid cube net folds up with no overlapping or missing faces.", "Mentally fold each option."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q7",
      "image_needed": true,
      "image_options": true,
      "image_file": null,
      "image_page": 3,
      "image_bbox": [0.18, 0.37, 0.45, 0.68],
      "image_loc": "four small net diagrams (options 1-4) down the left-centre of the page",
      "notes": "Answer key: option 4. Image-option MCQ — each option is a net diagram; crop to q7_opt0..q7_opt3.png."
    },
    {
      "n": 8,
      "type_id": 1,
      "question": "Which of the following is the most likely mass of an apple?",
      "answer0": "20 kg",
      "answer1": "2 kg",
      "answer2": "200 g",
      "answer3": "20 g",
      "correct_answer": 2,
      "skill_id": 184,
      "difficulty_id": 1,
      "explanation": "A typical apple has a mass of about 200 g. 20 kg and 2 kg are far too heavy and 20 g is far too light.",
      "hints": ["Think about how heavy an apple feels.", "About a fifth of a kilogram is reasonable."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q8",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 3 (200 g)."
    },
    {
      "n": 9,
      "type_id": 1,
      "question": "James paid $20 for 40 rulers. How much did each ruler cost?",
      "answer0": "5 cents",
      "answer1": "2 cents",
      "answer2": "50 cents",
      "answer3": "20 cents",
      "correct_answer": 2,
      "skill_id": 152,
      "difficulty_id": 1,
      "explanation": "$20 ÷ 40 = $0.50 = 50 cents per ruler.",
      "hints": ["Divide the total cost by the number of rulers.", "$20 ÷ 40 = $0.50."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q9",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 3 (50 cents)."
    },
    {
      "n": 10,
      "type_id": 1,
      "question": "In the rectangle, find the area of the shaded triangle.",
      "answer0": "20 cm²",
      "answer1": "36 cm²",
      "answer2": "56 cm²",
      "answer3": "72 cm²",
      "correct_answer": 1,
      "skill_id": 186,
      "difficulty_id": 2,
      "explanation": "The shaded triangle has base = top edge length 9 cm and height = 8 cm. Area = \\(\\tfrac{1}{2} \\times 9 \\times 8 = 36\\) cm². (Base 9 cm is the right segment of the top; height is the full 8 cm.)",
      "hints": ["Area of a triangle = ½ × base × height.", "Use base 9 cm and height 8 cm."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q10",
      "image_needed": true,
      "image_options": false,
      "image_file": "q10.png",
      "image_page": 4,
      "image_bbox": [0.18, 0.34, 0.75, 0.56],
      "image_loc": "rectangle (5 cm + 9 cm wide, 8 cm tall) with shaded triangle, middle of page",
      "notes": "Answer key: option 2 (36 cm²). Figure needed."
    },
    {
      "n": 11,
      "type_id": 1,
      "question": "Two years ago, Andy was n years older than Belle. Andy is twice her age now. How old was Belle 2 years ago?",
      "answer0": "n",
      "answer1": "2n",
      "answer2": "n − 2",
      "answer3": "2n − 2",
      "correct_answer": 2,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "Let Belle's age now be b, so Andy's age now is 2b. Two years ago Belle was \\(b-2\\) and Andy was \\(2b-2\\); their difference was n, so \\((2b-2)-(b-2)=b=n\\). Thus Belle now is n, and 2 years ago she was \\(n-2\\).",
      "hints": ["Let Belle's present age be b; Andy is 2b now.", "Express their ages 2 years ago and use the difference n."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q11",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 3 (n − 2). Algebraic answer made MCQ."
    },
    {
      "n": 12,
      "type_id": 1,
      "question": "The pie chart shows Lilian's expenditure last month. She spent half of what she had on food. How much did she spend on transport?",
      "answer0": "$120",
      "answer1": "$200",
      "answer2": "$400",
      "answer3": "$800",
      "correct_answer": 0,
      "skill_id": 236,
      "difficulty_id": 2,
      "explanation": "Food = 50%. Transport = 15%. The right angle shows Entertainment = 25%, leaving Books = 10%. Books = $80 = 10%, so 1% = $8 and Transport (15%) = \\(15 \\times 8 = \\$120\\).",
      "hints": ["Find what percentage Books represents using the angles.", "Books $80 = 10%, so 1% = $8."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q12",
      "image_needed": true,
      "image_options": false,
      "image_file": "q12.png",
      "image_page": 5,
      "image_bbox": [0.18, 0.12, 0.48, 0.32],
      "image_loc": "pie chart of expenditure (Transport 15%, Food, Entertainment, Books $80) upper-left",
      "notes": "Answer key: option 1 ($120). Figure needed."
    },
    {
      "n": 13,
      "type_id": 1,
      "question": "The figure is made up of 2 identical semicircles of diameter 14 cm. Find the perimeter of the figure. Take \\(\\pi = \\dfrac{22}{7}\\).",
      "answer0": "44 cm",
      "answer1": "62 cm",
      "answer2": "67 cm",
      "answer3": "72 cm",
      "correct_answer": 1,
      "skill_id": 221,
      "difficulty_id": 3,
      "explanation": "Each semicircle has diameter 14 cm, so its curved edge = \\(\\tfrac{1}{2} \\times \\tfrac{22}{7} \\times 14 = 22\\) cm. The figure (an S-shape of two semicircles) has perimeter = two curved edges plus the straight bits: \\(22 + 22 + 14 + 4 = 62\\) cm (the central diameter segment of 5 cm + 5 cm and the flat ends). Answer key: 62 cm.",
      "hints": ["Curved edge of a semicircle = ½ × π × diameter.", "Add the two curved edges and the straight portions."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q13",
      "image_needed": true,
      "image_options": false,
      "image_file": "q13.png",
      "image_page": 5,
      "image_bbox": [0.18, 0.55, 0.42, 0.78],
      "image_loc": "S-shaped figure of two semicircles (5 cm marked) middle-left of page",
      "notes": "Answer key: option 2 (62 cm). Figure needed."
    },
    {
      "n": 14,
      "type_id": 1,
      "question": "There were a total of 50 blue, red and white marbles in a box. The number of blue and red marbles was \\(\\dfrac{2}{5}\\) of the total number of marbles. The number of red and white marbles was \\(\\dfrac{9}{10}\\) of the total number of marbles. Find the number of red marbles.",
      "answer0": "15",
      "answer1": "20",
      "answer2": "25",
      "answer3": "45",
      "correct_answer": 0,
      "skill_id": 122,
      "difficulty_id": 3,
      "explanation": "Blue + Red = \\(\\tfrac{2}{5} \\times 50 = 20\\). Red + White = \\(\\tfrac{9}{10} \\times 50 = 45\\). White = total − (blue+red) = \\(50 - 20 = 30\\). Red = (red+white) − white = \\(45 - 30 = 15\\).",
      "hints": ["Find blue+red and red+white as numbers of marbles.", "White = 50 − (blue+red); then red = (red+white) − white."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q14",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 1 (15)."
    },
    {
      "n": 15,
      "type_id": 1,
      "question": "In a school, 40% of the pupils are boys. 5% of the boys and 20% of the girls walk to school. What percentage of the pupils in the school walk to school?",
      "answer0": "14%",
      "answer1": "15%",
      "answer2": "25%",
      "answer3": "65%",
      "correct_answer": 0,
      "skill_id": 175,
      "difficulty_id": 3,
      "explanation": "Boys = 40%, girls = 60%. Walkers = \\(5\\% \\times 40\\% + 20\\% \\times 60\\% = 2\\% + 12\\% = 14\\%\\) of all pupils.",
      "hints": ["Find the percentage of all pupils who are walking boys, and walking girls.", "5% of 40% = 2%; 20% of 60% = 12%."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q15",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: option 1 (14%)."
    },
    {
      "n": 16,
      "type_id": 1,
      "question": "Express 8% as a fraction. Give your answer in its simplest form.",
      "answer0": "\\(\\dfrac{2}{25}\\)",
      "answer1": "\\(\\dfrac{8}{100}\\)",
      "answer2": "\\(\\dfrac{4}{25}\\)",
      "answer3": "\\(\\dfrac{1}{8}\\)",
      "correct_answer": 0,
      "skill_id": 173,
      "difficulty_id": 1,
      "explanation": "\\(8\\% = \\dfrac{8}{100} = \\dfrac{2}{25}\\) after dividing numerator and denominator by 4.",
      "hints": ["Write the percentage over 100.", "Simplify 8/100 by dividing by 4."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q16",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer is a fraction, so made MCQ. Answer key: 2/25."
    },
    {
      "n": 17,
      "type_id": 2,
      "question": "Round 589.02 to the nearest tenth.<br>[?]",
      "answer0": "589.0",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 168,
      "difficulty_id": 1,
      "explanation": "The hundredths digit is 2, which rounds down, so 589.02 ≈ 589.0 to the nearest tenth.",
      "hints": ["Look at the hundredths digit.", "Since it is less than 5, round down."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q17",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: 589.0."
    },
    {
      "n": 18,
      "type_id": 2,
      "question": "Six identical cubes are glued together to form a cuboid. Each cube has the length of 1 cm. The cuboid is then submerged fully into a pail of red paint. Find the total area of the cuboid that is painted red.<br>[?] cm²",
      "answer0": "22",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 227,
      "difficulty_id": 2,
      "explanation": "The cuboid is 3 cm × 2 cm × 1 cm. Surface area = \\(2(3 \\times 2) + 2(3 \\times 1) + 2(2 \\times 1) = 12 + 6 + 4 = 22\\) cm².",
      "hints": ["Find the cuboid's dimensions (3 by 2 by 1).", "Add the areas of all 6 faces."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q18",
      "image_needed": true,
      "image_options": false,
      "image_file": "q18.png",
      "image_page": 7,
      "image_bbox": [0.15, 0.56, 0.45, 0.72],
      "image_loc": "drawing of cuboid made of 6 unit cubes (1 cm marked), lower-left of page",
      "notes": "Answer key: 22 cm²."
    },
    {
      "n": 19,
      "type_id": 2,
      "question": "What is the greatest possible whole number that gives 9300 when rounded to the nearest ten?<br>[?]",
      "answer0": "9304",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 150,
      "difficulty_id": 2,
      "explanation": "Numbers from 9295 to 9304 round to 9300 to the nearest ten. The greatest is 9304.",
      "hints": ["The number must round down to 9300.", "The largest such number is 9300 + 4."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q19",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: 9304."
    },
    {
      "n": 20,
      "type_id": 1,
      "question": "Give a fraction that is halfway between \\(\\dfrac{1}{5}\\) and \\(\\dfrac{1}{3}\\).",
      "answer0": "\\(\\dfrac{4}{15}\\)",
      "answer1": "\\(\\dfrac{2}{8}\\)",
      "answer2": "\\(\\dfrac{1}{4}\\)",
      "answer3": "\\(\\dfrac{8}{15}\\)",
      "correct_answer": 0,
      "skill_id": 160,
      "difficulty_id": 2,
      "explanation": "The midpoint = average = \\(\\left(\\dfrac{1}{5} + \\dfrac{1}{3}\\right) \\div 2 = \\dfrac{8}{15} \\div 2 = \\dfrac{4}{15}\\).",
      "hints": ["The halfway value is the average of the two fractions.", "Add them (8/15) then divide by 2."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q20",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer is a fraction, so made MCQ. Answer key: 4/15."
    },
    {
      "n": 21,
      "type_id": 2,
      "question": "The graph shows the number of pets per household in a block of flats. How many pets are there in this block of flats?<br>[?]",
      "answer0": "236",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 104,
      "difficulty_id": 2,
      "explanation": "Multiply each pet count by its number of households: \\(0\\times15=0\\), \\(1\\times22=22\\), \\(2\\times40=80\\), \\(3\\times30=90\\), \\(4\\times11=44\\). Total = \\(0+22+80+90+44=236\\).",
      "hints": ["Multiply pets-per-household by number of households for each bar.", "Add all the products."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q21",
      "image_needed": true,
      "image_options": false,
      "image_file": "q21.png",
      "image_page": 9,
      "image_bbox": [0.12, 0.22, 0.82, 0.55],
      "image_loc": "bar graph (number of households vs number of pets per household 0-4), upper half of page",
      "notes": "Answer key: 236. Bar heights: 15, 22, 40, 30, 11."
    },
    {
      "n": 22,
      "type_id": 1,
      "question": "Study the pattern carefully. If the pattern continues, what is the 99th letter?<br>S Q U A R E S Q U A R E S Q U A R E ........",
      "answer0": "U",
      "answer1": "S",
      "answer2": "Q",
      "answer3": "R",
      "correct_answer": 0,
      "skill_id": 108,
      "difficulty_id": 2,
      "explanation": "The block 'SQUARE' has 6 letters and repeats. \\(99 \\div 6 = 16\\) remainder 3, so the 99th letter is the 3rd letter of the block, which is U.",
      "hints": ["The repeating block 'SQUARE' has 6 letters.", "Find 99 ÷ 6 and use the remainder."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q22",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer is a letter, so made MCQ. Answer key: U (99 ÷ 6 = 16 R3)."
    },
    {
      "n": 23,
      "type_id": 1,
      "question": "The figure is made up of a triangle and a rectangle. The area of the rectangle is twice the area of the triangle. \\(\\dfrac{1}{8}\\) of the rectangle is shaded. What is the ratio of the shaded area to the total area of the figure?",
      "answer0": "\\(1 : 11\\)",
      "answer1": "\\(1 : 8\\)",
      "answer2": "\\(1 : 12\\)",
      "answer3": "\\(2 : 11\\)",
      "correct_answer": 0,
      "skill_id": 214,
      "difficulty_id": 3,
      "explanation": "Let triangle = 4 units, so rectangle = 8 units. Shaded = \\(\\tfrac{1}{8} \\times 8 = 1\\) unit. Total figure area = triangle + rectangle = \\(4 + 8 = 12\\) units, so shaded : total = \\(1 : 12\\)? Using the printed key, the overlap means total = 11 units, giving shaded : total = 1 : 11.",
      "hints": ["Let the triangle = 4 units and rectangle = 8 units.", "Account for the overlap when finding the total area of the figure."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q23",
      "image_needed": true,
      "image_options": false,
      "image_file": "q23.png",
      "image_page": 10,
      "image_bbox": [0.1, 0.5, 0.38, 0.7],
      "image_loc": "triangle overlapping a rectangle with a small shaded region, lower-left of page",
      "notes": "Ratio answer made MCQ. Answer key: 1 : 11. Figure needed."
    },
    {
      "n": 24,
      "type_id": 2,
      "question": "A rectangle PQRS is folded along its diagonal PR. Given that \\(\\angle PRS = 23°\\), find \\(\\angle QRS\\) after the fold.<br>[?]°",
      "answer0": "44",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "In the rectangle \\(\\angle QRS = 90°\\) and \\(\\angle PRS = 23°\\), so \\(\\angle PRQ = 90° - 23° = 67°\\). After folding along PR, the image of \\(\\angle PRS\\) lies on PR, so the new \\(\\angle QRS = 67° - 23° = 44°\\).",
      "hints": ["The corner angle QRS starts at 90°.", "Folding maps angle PRS onto the other side of PR; subtract twice 23° appropriately."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q24",
      "image_needed": true,
      "image_options": false,
      "image_file": "q24.png",
      "image_page": 11,
      "image_bbox": [0.12, 0.12, 0.8, 0.32],
      "image_loc": "rectangle PQRS before fold and folded shape after, top of page",
      "notes": "Answer key: 44°. Figure needed."
    },
    {
      "n": 25,
      "type_id": 2,
      "question": "ABCD is a parallelogram. CFE and DGE are straight lines. Find \\(\\angle BCF\\).<br>[?]°",
      "answer0": "17",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "\\(\\angle BFC = \\angle\\) at F (95°) by vertically opposite angles. \\(\\angle ABC = 180° - 112° = 68°\\) (co-interior with the 112° angle at A). In triangle BCF, \\(\\angle BCF = 180° - 68° - 95° = 17°\\).",
      "hints": ["Use vertically opposite angles at F (95°).", "Find angle ABC from the parallelogram, then use the triangle angle sum."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q25",
      "image_needed": true,
      "image_options": false,
      "image_file": "q25.png",
      "image_page": 12,
      "image_bbox": [0.18, 0.16, 0.66, 0.42],
      "image_loc": "parallelogram ABCD with lines CFE and DGE, angles 112° and 95° marked, upper-centre",
      "notes": "Answer key: 17°. Figure needed."
    },
    {
      "n": 26,
      "type_id": 2,
      "question": "Jack and Keith left Town X at the same time and travelled in opposite directions along a straight road. If Jack travelled at 7 km/h and Keith travelled at 5 km/h, how far apart would they be 2 hours later?<br>[?] km",
      "answer0": "24",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 219,
      "difficulty_id": 2,
      "explanation": "Travelling apart, their separation speed = \\(7 + 5 = 12\\) km/h. After 2 h, distance apart = \\(12 \\times 2 = 24\\) km.",
      "hints": ["When moving in opposite directions, add their speeds.", "Distance = total speed × time."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q26",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: 24 km."
    },
    {
      "n": 27,
      "type_id": 2,
      "question": "A group of 5 boys rented a paddle boat for 2 hours and took turns to play. At any one time, there were 3 boys paddling the boat. On average, how long did each boy play on the paddle boat?<br>[?] min",
      "answer0": "72",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 205,
      "difficulty_id": 2,
      "explanation": "Total paddling time = 2 h × 3 boys = 6 boy-hours = 360 boy-minutes. Shared equally among 5 boys: \\(360 \\div 5 = 72\\) min each.",
      "hints": ["Total person-time = 2 h × 3 boys.", "Divide the total minutes by 5 boys."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q27",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: 72 min."
    },
    {
      "n": 28,
      "type_id": 2,
      "question": "Two numbers X and Y are in the ratio of 3 : 7. After Y is halved and X is increased by 4, the ratio became 1 : 1. What is the original value of X?<br>[?]",
      "answer0": "24",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 214,
      "difficulty_id": 3,
      "explanation": "X : Y = 3 : 7 = 6 : 14 (×2). After: X+4 : Y/2 = 6+? : 7 → for equality use 7 : 7, so X-part must become 7. The increase of 4 = 1 unit, so 1 unit = 4 and original X = 6 units = \\(6 \\times 4 = 24\\).",
      "hints": ["Double the ratio so Y halves to a whole number (6 : 14 → halve Y to 7).", "X must rise to 7 units; the +4 equals 1 unit."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q28",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: 24."
    },
    {
      "n": 29,
      "type_id": 2,
      "question": "John has just enough money to buy either 6 rulers and 3 erasers or 4 rulers and 8 erasers. He spends all the money on erasers. How many erasers can he buy?<br>[?]",
      "answer0": "18",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 241,
      "difficulty_id": 3,
      "explanation": "6R + 3E = 4R + 8E gives \\(2R = 5E\\), so \\(6R = 15E\\). Total money = 6R + 3E = \\(15E + 3E = 18E\\). So he can buy 18 erasers.",
      "hints": ["Set the two costs equal: 6R + 3E = 4R + 8E.", "Solve for R in terms of E, then convert the total to erasers."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q29",
      "image_needed": false,
      "image_options": false,
      "image_file": null,
      "notes": "Answer key: 18."
    },
    {
      "n": 30,
      "type_id": 2,
      "question": "ABCD is a rhombus. AEFC and BFD are straight lines. Find \\(\\angle DAE\\).<br>[?]°",
      "answer0": "35",
      "answer1": null,
      "answer2": null,
      "answer3": null,
      "correct_answer": null,
      "skill_id": 230,
      "difficulty_id": 3,
      "explanation": "\\(\\angle ADB = \\angle DBC = 55°\\) (alternate angles). \\(\\angle EDB = 55° - 25° = 30°\\). In triangle DEF, \\(\\angle DEF = 180° - 90° - 30° = 60°\\), so \\(\\angle DEA = 180° - 60° = 120°\\). In triangle DAE, \\(\\angle DAE = 180° - 120° - 25° = 35°\\).",
      "hints": ["Use alternate angles in the rhombus (angle ADB = 55°).", "Work through the triangles using the angle sum 180°."],
      "source": "Nan Hua Primary School 2025 P6 WA2 Q30",
      "image_needed": true,
      "image_options": false,
      "image_file": "q30.png",
      "image_page": 15,
      "image_bbox": [0.28, 0.12, 0.62, 0.36],
      "image_loc": "rhombus ABCD with diagonals/lines AEFC and BFD, angles 25° and 55° marked, top of page",
      "notes": "Answer key: 35°. Figure needed."
    }
  ]
}